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(4x^2-2x/x^2+5x+4)/(2x/x^2+2x+1) how do i do this

 Jul 11, 2015

Best Answer 

 #1
avatar+118608 
+5

$$(4x^2-2x/x^2+5x+4)/(2x/x^2+2x+1)$$

 

I assume that you really mean

 

$$\\((4x^2-2x)/(x^2+5x+4))/((2x)/(x^2+2x+1) )\\\\
=\frac{4x^2-2x}{x^2+5x+4}\div\frac{2x}{x^2+2x+1}\\\\
=\frac{4x^2-2x}{x^2+5x+4}\times\frac{x^2+2x+1}{2x}\\\\
=\frac{x(x-2)}{(x+4)(x+1)}\times\frac{(x+1)(x+1)}{2x}\\\\
=\frac{(x-2)}{(x+4)}\times\frac{(x+1)}{2}\\\\
=\frac{(x-2)(x+1)}{2(x+4)}\\\\
=\frac{x^2-x-2}{2x+8}\\\\$$

.
 Jul 11, 2015
 #1
avatar+118608 
+5
Best Answer

$$(4x^2-2x/x^2+5x+4)/(2x/x^2+2x+1)$$

 

I assume that you really mean

 

$$\\((4x^2-2x)/(x^2+5x+4))/((2x)/(x^2+2x+1) )\\\\
=\frac{4x^2-2x}{x^2+5x+4}\div\frac{2x}{x^2+2x+1}\\\\
=\frac{4x^2-2x}{x^2+5x+4}\times\frac{x^2+2x+1}{2x}\\\\
=\frac{x(x-2)}{(x+4)(x+1)}\times\frac{(x+1)(x+1)}{2x}\\\\
=\frac{(x-2)}{(x+4)}\times\frac{(x+1)}{2}\\\\
=\frac{(x-2)(x+1)}{2(x+4)}\\\\
=\frac{x^2-x-2}{2x+8}\\\\$$

Melody Jul 11, 2015

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